# using a transformation to simplify a differential equation

• Jan 11th 2011, 07:56 PM
honda
using a transformation to simplify a differential equation
Hi all,
Is it just me or is there something strange going on here.

Consider the following differential equation:

$0=\frac{1}{2}-f(x)+f'(x)[3(\frac{1}{2}-x)-2(\frac{1}{2}-f(x))]$

Consider the transformation (x<1/2)

$h(x)=\frac{1/2-f(x)}{1/2-x}$.

Then the differential equation is supposed to become:

$h'(2h-x)(\frac{1}{2}-x)=2h(h-2)$

However, when I did it I got:

$h'(2h-3)(\frac{1}{2}-x)=2h(h-2)$.

I did the following:

Calculate f'(x):

$f(x)=\frac{1}{2}-h(x)(\frac{1}{2}-x)$
$=> f'(x)=h(x)-h'(x)(\frac{1}{2}-x)$

Substitute f'(x) into the DE, also replacing (1/2-f(x)), to get

$0=h(\frac{1}{2}-x)+(h-h'(\frac{1}{2}-x))[3(\frac{1}{2}-x)-2h(\frac{1}{2}-x)]$
$0=(\frac{1}{2}-x)[h+(h-h'(\frac{1}{2}-x))(3-2h)]$
$0=(\frac{1}{2}-x)[4h-2h^{2}-h'(\frac{1}{2}-x))(3-2h)]$
$h'(\frac{1}{2}-x)(3-2h)=2h(2-h)$
$h'(\frac{1}{2}-x)(2h-3)=2h(h-2)$

So what am I missing here (I'm 99% sure it's me who's wrong)?

Thanks very much!
• Jan 12th 2011, 02:33 AM
Sudharaka
Quote:

Originally Posted by honda
Hi all,
Is it just me or is there something strange going on here.

Consider the following differential equation:

$0=\frac{1}{2}-f(x)+f'(x)[3(\frac{1}{2}-x)-2(\frac{1}{2}-f(x))]$

Consider the transformation (x<1/2)

$h(x)=\frac{1/2-f(x)}{1/2-x}$.

Then the differential equation is supposed to become:

$h'(2h-x)(\frac{1}{2}-x)=2h(h-2)$

However, when I did it I got:

$h'(2h-3)(\frac{1}{2}-x)=2h(h-2)$.

I did the following:

Calculate f'(x):

$f(x)=\frac{1}{2}-h(x)(\frac{1}{2}-x)$
$=> f'(x)=h(x)-h'(x)(\frac{1}{2}-x)$

Substitute f'(x) into the DE, also replacing (1/2-f(x)), to get

$0=h(\frac{1}{2}-x)+(h-h'(\frac{1}{2}-x))[3(\frac{1}{2}-x)-2h(\frac{1}{2}-x)]$
$0=(\frac{1}{2}-x)[h+(h-h'(\frac{1}{2}-x))(3-2h)]$
$0=(\frac{1}{2}-x)[4h-2h^{2}-h'(\frac{1}{2}-x))(3-2h)]$
$h'(\frac{1}{2}-x)(3-2h)=2h(2-h)$
$h'(\frac{1}{2}-x)(2h-3)=2h(h-2)$

So what am I missing here (I'm 99% sure it's me who's wrong)?

Thanks very much!

Dear honda,